Worksheet Solutions: Pair of Linear Equations in Two Variables

# Pair of Linear Equations in Two Variables Class 10 Worksheet Maths Chapter 3

## Multiple Choice Questions

Q1: A pair of linear equations which have a unique solution x = 2, y = – 3 is:
(a) 2x – 3y = – 5, x + y = – 1
(b) 2x + 5y + 11 = 0, 4x + 10y + 22 = 0
(c) x – 4y – 14 = 0, 5x – y – 13 = 0
(d) 2x – y = 1, 3x + 2y = 0

Sol:  x – 4y – 14 = 0, 5x – y – 13 = 0

Q2: If a system of a pair of linear equations in two unknowns is consistent, then the lines representing the system will be
(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident

Sol: intersecting or coincident

Q3: The pair of equations x = 0 and y = 0 has
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

Sol: one solution

Q4: A pair of system of equations x = 2, y = -2; x = 3, y = – 3 when represented graphically enclose
(a) Square
(b) Trapezium
(c) Rectangle
(d) Triangle

Sol: Rectangle

Q5: If two lines are parallel to each other then the system of equations is
(a) consistent
(b) inconsistent
(c) consistent dependent
(d) (a) and (c) both

Sol: inconsistent

## Fill in the blanks

Q1: If in a system of equations corresponding to coefficients of member, equations are proportional then the system has ______________ solution (s).
Sol: infinite

Q2: A pair of linear equations is said to be inconsistent if its graph lines are ____________.
Sol: parallel
Q3:
A pair of linear equations is said to be ____________ if its graph lines intersect or coincide.
Sol: Consistent
Q4:
A consistent system of equations where straight lines fall on each other is also called _____________ system of equations.
Sol: Dependent
Q5: Solution of linear equations representing 2x – y = 0, 8x + y = 25 is ____________ .
Sol: x = 25,y = 5

Q1: In Fig., ABCD is a rectangle. Find the values of x and y.

Sol: Since ABCD is a rectangle
⇒ AB = CD and BC = AD
x + y = 30 …………….. (i)
x – y = 14 ……………. (ii)
(i) + (ii) ⇒ 2x = 44
⇒ x = 22
Plug in x = 22 in (i)
⇒ 22 + y = 30
⇒ y = 8

Q2: Name the geometrical figure enclosed by graph of the equations x + 7 = 0, y – 2 = 0 and x – 2 = 0, y + 7 = 0.
Sol: Clearly, a square of side 9 units is enclosed by lines.

Q3: If 51x + 23y = 116 and 23x + 51y = 106, then find the value of (x – y).
Sol: Given equations are:

51x + 23y = 116 ……….. (i)
23x + 51y = 106 ………… (ii)
Subtracting (ii) from (i)
28x – 28y = 10
28(x – y) = 10

⇒ (x – y) = = 514

Q4: For what value of V does the point (3, a) lie on the line represented by 2x – 3y = 5?
Sol: Since (3, a) lies on the equation 2x – 3y = 5.
∴ (3, a) must satisfy this equation
⇒ 2(3) – 3(a) = 5
⇒ 6 – 3a = 5
⇒ – 3a = 5 – 6 = – 1
a = 1/31
∴ a = 1/313

Q5: Determine whether the following system of linear equations is inconsistent or not.
3x – 5y = 20
6x – 10y = -40
Sol: Given
3x – 5y = 20 ……… (i)
6x – 10y = – 40 ………… (ii)

Hence given pair of linear equations are parallel.∴ It is inconsistent.

Q1: The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. Find the present ages of the son and the father.
Sol:
Let the father's age be y and the son's age be x.
First condition: y = 6x
Second condition: y + 4 = 4 (x + 4)
So, we have  6x + 4 = 4 x + 16
x = 6 and y = 36

Q2: If the lines x + 2y + 7 = 0 and 2x + ky + 18 = 0 intersect at a point, then find the value of k.
Sol: For a unique solution
Substituting the values

Q3:  Find the value of k for which the system of equations x + 2y -3 = 0 and ky + 5x + 7 = 0 has a unique solution.
Sol:
For a unique solution

Substituting the values

Q4: Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
2x + 3 y = 7
2αx + (a + b) y = 28
Sol:
We have

Solving this , we get a = 4 and b = 8

Q5: In a cyclic quadrilateral ABCD, Find the four angles.
a. ∠A = (2 x + 4), ∠B = (y + 3), ∠C = (2y + 10) , ∠D = (4x − 5) .
b. ∠A = (2 x − 1) , ∠ B = (y + 5) , ∠C = (2 y + 15) and ∠D = (4 x − 7)
Sol:
a. ∠ A = (2 x + 4) , ∠B = (y + 3), ∠C = (2 y + 10), ∠D = (4 x − 5)
In a cyclic quadrilateral, Opposite angles are supplementary.
∠A + ∠C = 180° and ∠ B + ∠ D = 180°
So, 2x + 4 + 2y + 10 = 180
or x + y = 83
y + 3 + 4 x − 5 = 180
or y + 4 x = 182
Solving the above equation by Substitution method
x = 33 and y = 50
So Angles are 700,530,1100,1270

b. ∠A = (2 x − 1) , ∠B = (y + 5) , ∠C = (2 y + 15) and ∠D = (4 x − 7)
Solving similarly, we get
650, 550, 1150, 1250

Q1: Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and the x-axis. Find the area of the shaded region.
Sol: The given system of equations is
2x + y – 6 = 0 ………. (i)
2x – y + 2 = 0 ……… (ii)
Let us write three solutions for each equation of the system in a table.
(i) ⇒ y = 6 – 2x
Table of solutions for 2x + y – 6 = 0

Similarly (ii) ⇒ y = 2x + 2
Table of solutions for 2x – y + 2 = 0

Plotting these points of each table of solutions on the same graph paper and joining them with a ruler, we obtain the graph of two lines represented by equations (i) and (ii) respectively as shown in the graph below. Since, the two lines intersect at point P(1, 4). Thus x = 1, y = 4 is the solution of the given system of equations. In the graph, the area is bounded by the lines and the x-axis is ∆PAB which is shaded. Draw PM ⊥ x-axis
Clearly, PM = y-coordinate of P(1, 4) = 4 units
Also, AB = 1 + 3 = 4 units
∴ Area of the shaded region = Area of ∆PAB = × AB × PM=  × 4 × 4 = 8 sq. units.

Q2: Draw the graphs of the following equations:
2x – y = 1, x + 2y = 13
(i) Find the solution of the equations from the graph.
(ii) Shade the triangular region formed by lines and the y-axis.

Sol: 2x – y = 1 …………….. (i)
x + 2y = 13 ………………. (ii)
Let us draw a table of values for (i) and (ii)
Plotting these points on the graph paper, we see that the two lines representing equations (i) and (ii) intersect at points (3, 5).
(i) Therefore, (3, 5) is the solution of a given system,
(ii) Also, two lines enclose a triangular region (∆ABC) with a y-axis shaded in the graph.

Q3: A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Sol: Let the speed of the train= x km/ hour
Speed of the car = y km/hour
Case I:
Total distance travelled = 370 km
Distance traveled by train = 250 km
Distance travelled by car = (370 – 250) km = 120 km

or 250y + 120x = 4xyor 60x + 125y = 2xy

Case II:
Total distance covered = 370 km
Distance covered by train = 130 km
Distance covered by car = (370 -130) km = 240
Time is taken by train = hour
Time is taken by car =  hour
According to 2nd condition,
= 4 hours 18 minutes
130x+240y=4310
1300y + 2400x = 43xy
or 2400x + 1300y = 43xy
Multiplying equation (i) by 40, we get

or x = 100
From (i) and (iii), we get 60(100) + 125y = 2(100)y
6000 + 125y = 200y
or (200 – 125) y = 6000
or 75y = 6000
or y = = 80
Hence, the speeds of the train and car is 100 km/hour and 80 km/hour respectively.

Q4: The taxi charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is ₹75 and for a journey of 15 km, the charge paid is ₹110. What will a person have to pay for traveling a distance of 25 km?
Sol: Let fixed charges for taxi be ₹x and charges for covering distance be ₹y per km.
Then, according to the question, we have
x + 10y = 75 ……… (i)
and x + 15y = 110
Subtracting (i) from (ii), we get
5y = 35 ⇒ y = 35 ÷ 5 = 7
Putting y = 7 in (i), we get
x + 10 (7) = 75
⇒ x = 75 – 70 = 5
∴ The person will have to pay for travelling a distance of 25 km = x + 25y = 5 + 25(7) = ₹180.

Q5: Solve the following system by drawing their graph:
(3/2)x – (5/4)y = 6, 6x – 6y = 20.
Determine whether these are consistent, inconsistent, or dependent.

Sol:

Plotting the points and joining by a ruler in each case. Here, we see that the graph of given equations are parallel lines. The two lines have no point in common. The given system of equations has no solution and is, therefore, inconsistent.

The document Pair of Linear Equations in Two Variables Class 10 Worksheet Maths Chapter 3 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Pair of Linear Equations in Two Variables Class 10 Worksheet Maths Chapter 3

 1. What is the concept of pair of linear equations in two variables?
Ans. Pair of linear equations in two variables refers to a set of two equations that involve two variables, typically represented as x and y. The goal is to find the values of x and y that satisfy both equations simultaneously.
 2. How can we solve a pair of linear equations in two variables?
Ans. There are multiple methods to solve a pair of linear equations in two variables. One common method is the substitution method, where one equation is solved for one variable and then substituted into the other equation. The resulting equation with only one variable can then be solved to find its value, and subsequently, the value of the other variable. Another method is the elimination method, where the two equations are manipulated such that when added or subtracted, one variable gets eliminated, resulting in a new equation with only one variable. This equation can then be solved to find its value, and the value can be substituted back into one of the original equations to find the value of the other variable.
 3. What is the difference between consistent and inconsistent pair of linear equations?
Ans. A pair of linear equations is said to be consistent if it has at least one solution, which means the two equations intersect at a point. This implies that the equations are not parallel or coincident lines. On the other hand, a pair of linear equations is considered inconsistent if it has no solution, which means the two equations are parallel and do not intersect.
 4. Can a pair of linear equations have infinitely many solutions?
Ans. Yes, a pair of linear equations can have infinitely many solutions. This occurs when the two equations represent the same line, meaning they are coincident. In this case, any point on the line satisfies both equations, resulting in infinitely many solutions.
 5. How can pair of linear equations be applied in real-life situations?
Ans. Pair of linear equations can be applied in various real-life situations. For example, they can be used to solve problems related to cost and revenue, such as finding the number of units to be sold to break even or maximize profit. They can also be used in problems involving speed and distance, such as finding the time taken to travel a certain distance at different speeds. Additionally, pair of linear equations can be used in problems related to mixtures, percentages, and ratios, among others.

## Mathematics (Maths) Class 10

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